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Chapter 15: Problem 41

A \(3.00-\mathrm{m}\) -long string, fixed at both ends, has a mass of \(6.00 \mathrm{~g}\). If you want to set up a standing wave in this string having a frequency of \(300 . \mathrm{Hz}\) and three antinodes, what tension should you put the string under?

Short Answer

Expert verified
To find the tension required in the string, follow these steps: 1. Calculate the linear mass density: μ = 0.00200 kg/m. 2. Calculate the period: T = 0.00333 s. 3. Determine the wavelength: λ = 2.00 m. 4. Calculate the wave velocity: v = 600 m/s. 5. Calculate the tension: T = 720 N. The tension required is 720 N.

Step by step solution

01

Determine the linear mass density and period

First, we'll calculate the linear mass density \(μ\) by dividing the mass of the string by its length: $$ μ = \frac{m}{L} = \frac{6.00\,\text{g}}{3.00\,\text{m}} = 2.00\,\text{g/m} = 0.00200\,\text{kg/m} $$ Next, we'll find the period \(T\) using the given frequency \(f\): $$ T = \frac{1}{f} = \frac{1}{300\,\text{Hz}} = 0.00333\,\text{s} $$
02

Find the wavelength

Since we have 3 antinodes in the string, it means we have \(1.5\) complete wavelengths over the length of the string. Therefore, we can find the wavelength \(λ\) by dividing the length of the string by \(1.5\): $$ λ = \frac{L}{1.5} = \frac{3.00\,\text{m}}{1.5} = 2.00\,\text{m} $$
03

Calculate the wave velocity

Now we have the wavelength \(λ\) and the period \(T\), we can find the wave velocity \(v\) using the wave equation: $$ v = \frac{λ}{T} = \frac{2.00\,\text{m}}{0.00333\,\text{s}} = 600\,\text{m/s} $$
04

Calculate the tension

Finally, we can use the formula \(v = \sqrt{\frac{T}{μ}}\) to determine the required tension in the string to create the desired standing wave: $$ T = μv^2 = (0.00200\,\text{kg/m})(600\,\text{m/s})^2 = 720\,\text{N} $$ The tension required to create the desired standing wave with a frequency of \(300\,\text{Hz}\) and three antinodes in the string is \(720\,\text{N}\).

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Most popular questions from this chapter

A string is made to oscillate, and a standing wave with three antinodes is created. If the tension in the string is increased by a factor of 4 a) the number of antinodes increases. b) the number of antinodes remains the same. c) the number of antinodes decreases. d) the number of antinodes will equal the number of nodes.

The speed of light waves in air is greater than the speed of sound in air by about a factor of a million. Given a sound wave and a light wave of the same wavelength, both traveling through air, which statement about their frequencies is true? a) The frequency of the sound wave will be about a million times greater than that of the light wave. b) The frequency of the sound wave will be about a thousand times greater than that of the light wave. c) The frequency of the light wave will be about a thousand times greater than that of the sound wave. d) The frequency of the light wave will be about a million times greater than that of the sound wave. e) There is insufficient information to determine the relationship between the two frequencies.

A guitar string with a mass of \(10.0 \mathrm{~g}\) is \(1.00 \mathrm{~m}\) long and attached to the guitar at two points separated by \(65.0 \mathrm{~cm} .\) a) What is the frequency of the first harmonic of this string when it is placed under a tension of \(81.0 \mathrm{~N} ?\) b) If the guitar string is replaced by a heavier one that has a mass of \(16.0 \mathrm{~g}\) and is \(1.00 \mathrm{~m}\) long, what is the frequency of the replacement string's first harmonic?

The tension in a 2.7 -m-long, 1.0 -cm-diameter steel cable \(\left(\rho=7800 \mathrm{~kg} / \mathrm{m}^{3}\right)\) is \(840 \mathrm{~N}\). What is the fundamental frequency of vibration of the cable?

A small ball floats in the center of a circular pool that has a radius of \(5.00 \mathrm{~m}\). Three wave generators are placed at the edge of the pool, separated by \(120 .\). The first wave generator operates at a frequency of \(2.00 \mathrm{~Hz}\). The second wave generator operates at a frequency of \(3.00 \mathrm{~Hz}\). The third wave generator operates at a frequency of \(4.00 \mathrm{~Hz}\). If the speed of each water wave is \(5.00 \mathrm{~m} / \mathrm{s}\), and the amplitude of the waves is the same, sketch the height of the ball as a function of time from \(t=0\) to \(t=2.00 \mathrm{~s}\), assuming that the water surface is at zero height. Assume that all the wave generators impart a phase shift of zero. How would your answer change if one of the wave generators was moved to a different location at the edge of the pool?
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